Ta có: \(\frac{x-2019}{2018}+\frac{x-2018}{2017}=\frac{x-2017}{2016}+\frac{x-2016}{2015}\)

\(\Leftrightarrow\left(\frac{x-2019}{2018}+1\right)+\left(\frac{x-2018}{2017}+1\right)=\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2016}{2015}+1\right)\)

\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}=\frac{x-1}{2016}+\frac{x-1}{2015}\)

\(\Leftrightarrow\frac{x-1}{2018}+\frac{x-1}{2017}-\frac{x-1}{2016}-\frac{x-1}{2015}=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)

\(\Leftrightarrow x-1=0\)( vì \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\))

\(\Leftrightarrow x=1\)

Vạy x=1

trừ mỗi vế cho 2 rồi tách -2 thành -1và -1

X=1 nhé

     \(\frac{x+2014}{2015}+\frac{x+2015}{2016}=\frac{x+2016}{2017}+\frac{x+2017}{2018}\)

\(\Leftrightarrow\)\(\frac{x+2014}{2015}-1+\frac{x+2015}{2016}-1=\frac{x+2016}{2017}-1+\frac{x+2017}{2018}-1\)

\(\Leftrightarrow\)\(\frac{x-1}{2015}+\frac{x-1}{2016}=\frac{x-1}{2017}+\frac{x-1}{2018}\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)

\(\Leftrightarrow\)\(x-1=0\)   ( do  1/2015 + 1/2016 - 1/2017 - 1/2018 # 0 )

\(\Leftrightarrow\) \(x=1\)

Tìm x biết:

\(\frac{x}{2018}+\frac{x+1}{2017}+\frac{x+2}{2016}+\frac{x+3}{2015}=-4\)

Giải:Ta có:\(\frac{x}{2018}+\frac{x+1}{2017}+\frac{x+2}{2016}+\frac{x+3}{2015}=-4\)

\(\Rightarrow\frac{x}{2018}+1+\frac{x+1}{2017}+1+\frac{x+2}{2016}+1+\frac{x+3}{2015}+1=0\)

\(\Rightarrow\frac{x+2018}{2018}+\frac{x+2018}{2017}+\frac{x+2018}{2016}+\frac{x+2018}{2015}=0\)

\(\Rightarrow\left(x+2018\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}\right)=0\)

\(\Rightarrow x+2018=0\) vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}>0\)

\(\Rightarrow x=-2018\)

Vậy x=-2018 thỏa mãn

x2018 +x+12017 +x+22016 +x+32015 =−4

⇒x2018 +1+x+12017 +1+x+22016 +1+x+32015 +1=0

⇒x+20182018 +x+20182017 +x+20182016 +x+20182015 =0

⇒(x+2018)(12018 +12017 +12016 +12015 )=0

⇒x+2018=0 vì 12018 +12017 +12016 +12015 >0

⇒x=−2018

Vậy x=-2018 thỏa mãn

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

a) \(\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}=\frac{x+1}{8}+\frac{x+1}{9}\)

\(\Leftrightarrow\frac{x+1}{5}+\frac{x+1}{6}+\frac{x+1}{7}-\frac{x+1}{8}-\frac{x+1}{9}=0\)

\(\Leftrightarrow\left(x+1\right).\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{8}-\frac{1}{9}\right)=0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=0-1\)

\(\Rightarrow x=-1\)

Vậy \(x=-1.\)

Mình chỉ làm câu a) thôi nhé.

Chúc bạn học tốt!

Lê Thị Thục HiềnTrần Thanh PhươngVũ Minh Tuấn?Amanda?Nguyễn Việt LâmHISINOMA KINIMADONguyễn Huy TúAkai Haruma

\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\frac{x+2015}{5}+1+\frac{x+2016}{4}+1=\frac{x+2017}{3}+1+\frac{x+2018}{2}+1\)

\(\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)

\(\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\)

\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

mà \(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)nên

\(x+2020=0\)

\(x=-2020\)

Cộng 1 vào 2 vế ta có

\(\frac{x+2015}{5}+1+\frac{x+2016}{4}+1=\frac{x+2017}{3}+1+\frac{x+2018}{2}+1\)

\(\left(\frac{x+2015}{5}+\frac{5}{5}\right)+\left(\frac{x+2016}{4}+\frac{4}{4}\right)=\left(\frac{x+2017}{3}+\frac{3}{3}\right)+\left(\frac{x+2018}{2}+\frac{2}{2}\right)\)

\(\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\)

\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

Vì \(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)

nên \(x+2020=0\Rightarrow x=-2020\)

help me

\(a)\) Ta có : 

\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)

\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)

\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)

\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)

Lại có : 

\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)

\(\Rightarrow\)\(x=2019\)

Vậy \(x=2019\)

Chúc bạn học tốt ~ 

\(b)\) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(1-\frac{2}{x+1}=\frac{2017}{2019}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=1-\frac{2017}{2019}\)

\(\Leftrightarrow\)\(\frac{2}{x+1}=\frac{2}{2019}\)

\(\Leftrightarrow\)\(x+1=2019\)

\(\Leftrightarrow\)\(x=2019-1\)

\(\Leftrightarrow\)\(x=2018\)

Vậy \(x=2018\)

Chúc bạn học tốt ~ 

Đáp án là : 

Tìm x : 

x = -2019 

x= -2019

\(\frac{x+2}{2018}+\frac{x+3}{2017}+\frac{x+4}{2016}=-3\)

\(\frac{x+2}{2018}+1+\frac{x+3}{2017}+1+\frac{x+4}{2016}+1=0\)

\(\frac{x+2+2018}{2018}+\frac{x+3+2017}{2017}+\frac{x+4+2016}{2016}=0\)

\(\frac{x+2020}{2018}+\frac{x+2020}{2017}+\frac{x+2020}{2016}=0\)

\(\left(x+2020\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)

\(\Rightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

#Sakura

\(\frac{x+2}{2018}+\frac{x+3}{2017}+\frac{x+4}{2016}=-\overrightarrow{3}\)

=>\(\frac{x+2}{2018}+1+\frac{x+3}{2017}+1+\frac{x+4}{2016}+1=0\)

=>\(\frac{x+2020}{2018}+\frac{x+2020}{2017}+\frac{x+2020}{2016}=0\)

=>\(\left(x+2020\right):\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)

=>\(\left(x+2020\right)=0\)

=>\(x=0-2020\)

=>\(x=-2020\)

vậy ....

chúc bạn học tốt!

x-2/2016 + x-3/2017 + x-4/2018 +3=0

a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2002}{2}\)

\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

Vậy : \(x=-2020\)

Chúc bạn học tốt !!

a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\\ \left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\\ \frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\\ \frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\\ \left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\\ \Rightarrow x+2020=0\\ \Rightarrow x=-2020\)

Vậy x = -2020

b) \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\\ \left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)=\left(\frac{x+2017}{7}-1\right)+\left(\frac{x+2018}{8}-1\right)\\ \frac{x+2010}{5}+\frac{x+2010}{6}=\frac{x+2010}{7}+\frac{x+2010}{8}\\ \frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\\ \left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)

Vậy x = -2010